3.1.94 \(\int \frac {x^{7/2}}{\sqrt {b x+c x^2}} \, dx\) [94]

Optimal. Leaf size=108 \[ -\frac {32 b^3 \sqrt {b x+c x^2}}{35 c^4 \sqrt {x}}+\frac {16 b^2 \sqrt {x} \sqrt {b x+c x^2}}{35 c^3}-\frac {12 b x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c} \]

[Out]

-12/35*b*x^(3/2)*(c*x^2+b*x)^(1/2)/c^2+2/7*x^(5/2)*(c*x^2+b*x)^(1/2)/c-32/35*b^3*(c*x^2+b*x)^(1/2)/c^4/x^(1/2)
+16/35*b^2*x^(1/2)*(c*x^2+b*x)^(1/2)/c^3

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \begin {gather*} -\frac {32 b^3 \sqrt {b x+c x^2}}{35 c^4 \sqrt {x}}+\frac {16 b^2 \sqrt {x} \sqrt {b x+c x^2}}{35 c^3}-\frac {12 b x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/Sqrt[b*x + c*x^2],x]

[Out]

(-32*b^3*Sqrt[b*x + c*x^2])/(35*c^4*Sqrt[x]) + (16*b^2*Sqrt[x]*Sqrt[b*x + c*x^2])/(35*c^3) - (12*b*x^(3/2)*Sqr
t[b*x + c*x^2])/(35*c^2) + (2*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c)

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rubi steps

\begin {align*} \int \frac {x^{7/2}}{\sqrt {b x+c x^2}} \, dx &=\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {(6 b) \int \frac {x^{5/2}}{\sqrt {b x+c x^2}} \, dx}{7 c}\\ &=-\frac {12 b x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}+\frac {\left (24 b^2\right ) \int \frac {x^{3/2}}{\sqrt {b x+c x^2}} \, dx}{35 c^2}\\ &=\frac {16 b^2 \sqrt {x} \sqrt {b x+c x^2}}{35 c^3}-\frac {12 b x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {\left (16 b^3\right ) \int \frac {\sqrt {x}}{\sqrt {b x+c x^2}} \, dx}{35 c^3}\\ &=-\frac {32 b^3 \sqrt {b x+c x^2}}{35 c^4 \sqrt {x}}+\frac {16 b^2 \sqrt {x} \sqrt {b x+c x^2}}{35 c^3}-\frac {12 b x^{3/2} \sqrt {b x+c x^2}}{35 c^2}+\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.49 \begin {gather*} \frac {2 \sqrt {x (b+c x)} \left (-16 b^3+8 b^2 c x-6 b c^2 x^2+5 c^3 x^3\right )}{35 c^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/Sqrt[b*x + c*x^2],x]

[Out]

(2*Sqrt[x*(b + c*x)]*(-16*b^3 + 8*b^2*c*x - 6*b*c^2*x^2 + 5*c^3*x^3))/(35*c^4*Sqrt[x])

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Maple [A]
time = 0.41, size = 48, normalized size = 0.44

method	result	size
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-5 c^{3} x^{3}+6 b \,c^{2} x^{2}-8 b^{2} c x +16 b^{3}\right )}{35 \sqrt {x}\, c^{4}}\)	\(48\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-5 c^{3} x^{3}+6 b \,c^{2} x^{2}-8 b^{2} c x +16 b^{3}\right )}{35 \sqrt {x \left (c x +b \right )}\, c^{4}}\)	\(53\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-5 c^{3} x^{3}+6 b \,c^{2} x^{2}-8 b^{2} c x +16 b^{3}\right ) \sqrt {x}}{35 c^{4} \sqrt {c \,x^{2}+b x}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/35/x^(1/2)*(x*(c*x+b))^(1/2)*(-5*c^3*x^3+6*b*c^2*x^2-8*b^2*c*x+16*b^3)/c^4

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Maxima [A]
time = 0.27, size = 53, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (5 \, c^{4} x^{4} - b c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 8 \, b^{3} c x - 16 \, b^{4}\right )}}{35 \, \sqrt {c x + b} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*c^4*x^4 - b*c^3*x^3 + 2*b^2*c^2*x^2 - 8*b^3*c*x - 16*b^4)/(sqrt(c*x + b)*c^4)

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Fricas [A]
time = 1.62, size = 49, normalized size = 0.45 \begin {gather*} \frac {2 \, {\left (5 \, c^{3} x^{3} - 6 \, b c^{2} x^{2} + 8 \, b^{2} c x - 16 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{35 \, c^{4} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*c^3*x^3 - 6*b*c^2*x^2 + 8*b^2*c*x - 16*b^3)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {7}{2}}}{\sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**(7/2)/sqrt(x*(b + c*x)), x)

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Giac [A]
time = 2.00, size = 61, normalized size = 0.56 \begin {gather*} -\frac {2 \, \sqrt {c x + b} b^{3}}{c^{4}} + \frac {32 \, b^{\frac {7}{2}}}{35 \, c^{4}} + \frac {2 \, {\left (5 \, {\left (c x + b\right )}^{\frac {7}{2}} - 21 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )}}{35 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-2*sqrt(c*x + b)*b^3/c^4 + 32/35*b^(7/2)/c^4 + 2/35*(5*(c*x + b)^(7/2) - 21*(c*x + b)^(5/2)*b + 35*(c*x + b)^(
3/2)*b^2)/c^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{7/2}}{\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(7/2)/(b*x + c*x^2)^(1/2), x)

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